Question:
A spherical balloon loses its volume due to escape of air from it in such a way that decrease of volume at any instant is proportional to its surface area. Show that the radius is decreasing at a constant rate.
A spherical balloon loses its volume due to escape of air from it in such a way that decrease of volume at any instant is proportional to its surface area. Show that the radius is decreasing at a constant rate.
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In rate of change problems involving spheres, notice that the derivative of volume with respect to radius (\(dV/dr\)) is exactly the surface area.
Always distinguish between "rate of change" (derivative) and "rate of decrease" (negative of the derivative).
Always distinguish between "rate of change" (derivative) and "rate of decrease" (negative of the derivative).
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Volume of a sphere: \( V = \frac{4}{3}\pi r^3 \).
• Surface area of a sphere: \( S = 4\pi r^2 \).
• Rate of change: The derivative with respect to time \( t \).
• Proportionality: \( y \propto x \Rightarrow y = kx \), where \( k \) is a constant.
Step 1: Relate the rate of change of volume to surface area
Let \( V \) be the volume and \( S \) be the surface area of the spherical balloon at time \( t \). According to the problem, the rate of decrease of volume is proportional to the surface area: \[ -\frac{dV}{dt} \propto S \] \[ -\frac{dV}{dt} = kS \] where \( k \) is a positive constant of proportionality.
Step 2: Differentiate the volume formula with respect to time
We know that \( V = \frac{4}{3}\pi r^3 \). Differentiating both sides with respect to time \( t \) using the chain rule: \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) \] \[ \frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} \] \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Step 3: Substitute expressions for surface area and volume rate into the proportionality equation
Substituting \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \) and \( S = 4\pi r^2 \) into the equation from
Step 1: \[ -(4\pi r^2 \frac{dr}{dt}) = k(4\pi r^2) \] Dividing both sides by \( 4\pi r^2 \) (since \( r \neq 0 \)): \[ -\frac{dr}{dt} = k \] \[ \frac{dr}{dt} = -k \]
Step 4: Conclusion
The rate of change of the radius \( \frac{dr}{dt} \) is a constant (\(-k\)). The negative sign indicates that the radius is decreasing. Since \( k \) is a constant, the radius is decreasing at a constant rate.
• Volume of a sphere: \( V = \frac{4}{3}\pi r^3 \).
• Surface area of a sphere: \( S = 4\pi r^2 \).
• Rate of change: The derivative with respect to time \( t \).
• Proportionality: \( y \propto x \Rightarrow y = kx \), where \( k \) is a constant.
Step 1: Relate the rate of change of volume to surface area
Let \( V \) be the volume and \( S \) be the surface area of the spherical balloon at time \( t \). According to the problem, the rate of decrease of volume is proportional to the surface area: \[ -\frac{dV}{dt} \propto S \] \[ -\frac{dV}{dt} = kS \] where \( k \) is a positive constant of proportionality.
Step 2: Differentiate the volume formula with respect to time
We know that \( V = \frac{4}{3}\pi r^3 \). Differentiating both sides with respect to time \( t \) using the chain rule: \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) \] \[ \frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} \] \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]
Step 3: Substitute expressions for surface area and volume rate into the proportionality equation
Substituting \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \) and \( S = 4\pi r^2 \) into the equation from
Step 1: \[ -(4\pi r^2 \frac{dr}{dt}) = k(4\pi r^2) \] Dividing both sides by \( 4\pi r^2 \) (since \( r \neq 0 \)): \[ -\frac{dr}{dt} = k \] \[ \frac{dr}{dt} = -k \]
Step 4: Conclusion
The rate of change of the radius \( \frac{dr}{dt} \) is a constant (\(-k\)). The negative sign indicates that the radius is decreasing. Since \( k \) is a constant, the radius is decreasing at a constant rate.
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